Expert: Ahmed Salami - 11/17/2008

Question
f(x)= ln(sin^-1(2x))
a) find domain of f
b) find f'(x)

Answer
Hi Lauren,
f(x) = ln [sin^-1 (2x)]
a) For the domain of f(x), sin^-1 (2x) needs to be positive because of the ln function. But sin^-1 (2x) ranges from -π/2 to π/2, we take the positive interval 0 to π/2. Therefore, 2x lies in the interval
0 < 2x < 1
i.e
0 < x < 1/2

b) Let sin^-1 (2x) = y
f(x) = lny
f'(x) = (d/dx)f(x)
     = (d/dy)f(x) . dy/dx
     = (d/dy)lny . dy/dx
but (d/dy)lny = 1/y
             = 1/[sin^-1 (2x)]
dy/dx = 1/[sqrt(1 - (2x)^2)]
         = 1/[sqrt(1 - 4x^2)]
Therefore,
f'(x) = 1/[sin^-1 (2x)].1/[sqrt(1 - 4x^2)]
     = 1/[sin^-1 (2x)][sqrt(1 - 4x^2)]

Note that (d/dp)lnp = 1/p
and
(d/dp)sin^-1 (p) = 1/sqrt(1 - p^2)

Please dont be confused, i know that trigonometry can be cumbersome and ambiguous at times and can even be considered in a wider manner but you just need to take your time to make careful considerations and you'll be fine.

Regards.